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Circles Abound (2) (Posted on 2004-10-17) Difficulty: 4 of 5
I've a black circle of radius 5.

I wish to create 5 identical white circles of some lesser radius, which I will place on top of the black circle and completely obscure (cover) the black circle.

What is the smallest radius which I can make the smaller circles and still meet my requirement?

No Solution Yet Submitted by SilverKnight    
Rating: 3.3333 (3 votes)

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Sol. Comment 11 of 11 |

Five identical smallest circles which cover the black circles equally spaced around the black circle would meet at five points (p1, p2, p3, p4, p5) on the black circle spaced equally forming a regular pentagon each with sides of length 5.  The radius of the smaller circles then can be found by

1)  extracting the regular triangle (with the length of sides 5 each) formed by two points on the black circles and the center of the black circle and

2) finding the gravity point of the triangle and

3) finding the length of the gravity point to the vertex of the triangle

Let r = the radius of the smaller circle = the length from the gravity point to any of the vertices of the triangle.

Then r + r/2 is the height of the triangle. Now, consider the smaller right triangle inside this regular triangle with sides r and r/2  and 2.5 (half of the 5 which is the side of the regular triangle).  Find r by pythagorean triangle.   

r^2 = (r/2)^2 + (5/2)^2

r^2 - (r/2)^2 = 25/4

3r^2/4 = 25

r^2 = 25/3

r =square root of (25/3) = 5 times square root of 3/3

approximately 2.8867513

  Posted by doremi on 2004-10-25 21:07:02
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